3.1866 \(\int \frac {(1-2 x)^{3/2} (3+5 x)}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=81 \[ \frac {(1-2 x)^{5/2}}{42 (3 x+2)^2}-\frac {71 (1-2 x)^{3/2}}{126 (3 x+2)}-\frac {71}{63} \sqrt {1-2 x}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}} \]

[Out]

1/42*(1-2*x)^(5/2)/(2+3*x)^2-71/126*(1-2*x)^(3/2)/(2+3*x)+71/189*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-
71/63*(1-2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {78, 47, 50, 63, 206} \[ \frac {(1-2 x)^{5/2}}{42 (3 x+2)^2}-\frac {71 (1-2 x)^{3/2}}{126 (3 x+2)}-\frac {71}{63} \sqrt {1-2 x}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)^(3/2)*(3 + 5*x))/(2 + 3*x)^3,x]

[Out]

(-71*Sqrt[1 - 2*x])/63 + (1 - 2*x)^(5/2)/(42*(2 + 3*x)^2) - (71*(1 - 2*x)^(3/2))/(126*(2 + 3*x)) + (71*ArcTanh
[Sqrt[3/7]*Sqrt[1 - 2*x]])/(9*Sqrt[21])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2} (3+5 x)}{(2+3 x)^3} \, dx &=\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}+\frac {71}{42} \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2} \, dx\\ &=\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}-\frac {71}{42} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}-\frac {71}{18} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}+\frac {71}{18} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {71}{63} \sqrt {1-2 x}+\frac {(1-2 x)^{5/2}}{42 (2+3 x)^2}-\frac {71 (1-2 x)^{3/2}}{126 (2+3 x)}+\frac {71 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 48, normalized size = 0.59 \[ \frac {(1-2 x)^{5/2} \left (245-284 (3 x+2)^2 \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {3}{7}-\frac {6 x}{7}\right )\right )}{10290 (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)^(3/2)*(3 + 5*x))/(2 + 3*x)^3,x]

[Out]

((1 - 2*x)^(5/2)*(245 - 284*(2 + 3*x)^2*Hypergeometric2F1[2, 5/2, 7/2, 3/7 - (6*x)/7]))/(10290*(2 + 3*x)^2)

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fricas [A]  time = 1.08, size = 75, normalized size = 0.93 \[ \frac {71 \, \sqrt {21} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (120 \, x^{2} + 235 \, x + 101\right )} \sqrt {-2 \, x + 1}}{378 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="fricas")

[Out]

1/378*(71*sqrt(21)*(9*x^2 + 12*x + 4)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(120*x^2 + 235*x
+ 101)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [A]  time = 1.24, size = 77, normalized size = 0.95 \[ -\frac {71}{378} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} + \frac {225 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 511 \, \sqrt {-2 \, x + 1}}{108 \, {\left (3 \, x + 2\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="giac")

[Out]

-71/378*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 20/27*sqrt(-2*x
+ 1) + 1/108*(225*(-2*x + 1)^(3/2) - 511*sqrt(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 57, normalized size = 0.70 \[ \frac {71 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{189}-\frac {20 \sqrt {-2 x +1}}{27}-\frac {4 \left (-\frac {25 \left (-2 x +1\right )^{\frac {3}{2}}}{4}+\frac {511 \sqrt {-2 x +1}}{36}\right )}{3 \left (-6 x -4\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(3/2)*(5*x+3)/(3*x+2)^3,x)

[Out]

-20/27*(-2*x+1)^(1/2)-4/3*(-25/4*(-2*x+1)^(3/2)+511/36*(-2*x+1)^(1/2))/(-6*x-4)^2+71/189*arctanh(1/7*21^(1/2)*
(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.22, size = 83, normalized size = 1.02 \[ -\frac {71}{378} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {20}{27} \, \sqrt {-2 \, x + 1} + \frac {225 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 511 \, \sqrt {-2 \, x + 1}}{27 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)/(2+3*x)^3,x, algorithm="maxima")

[Out]

-71/378*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 20/27*sqrt(-2*x + 1) + 1/
27*(225*(-2*x + 1)^(3/2) - 511*sqrt(-2*x + 1))/(9*(2*x - 1)^2 + 84*x + 7)

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mupad [B]  time = 0.06, size = 63, normalized size = 0.78 \[ \frac {71\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{189}-\frac {20\,\sqrt {1-2\,x}}{27}-\frac {\frac {511\,\sqrt {1-2\,x}}{243}-\frac {25\,{\left (1-2\,x\right )}^{3/2}}{27}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(3/2)*(5*x + 3))/(3*x + 2)^3,x)

[Out]

(71*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/189 - (20*(1 - 2*x)^(1/2))/27 - ((511*(1 - 2*x)^(1/2))/243 -
(25*(1 - 2*x)^(3/2))/27)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(3/2)*(3+5*x)/(2+3*x)**3,x)

[Out]

Timed out

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